CUET · MATHS · PYQ PAPER 2023
The value of integral \(\int \sqrt{4 x^2+9} d x\) is
- A \(\frac{x}{2} \sqrt{4 x^2+9}+\frac{9}{2} \log \left|2 x+\sqrt{4 x^2+9}\right|+C\)
- B \(\frac{x}{2} \sqrt{4 x^2+9}+\frac{3}{2} \log \left|2 x+\sqrt{4 x^2+9}\right|+C\)
- C \(2 x \sqrt{4 x^2+9}+\frac{9}{2} \log \left|2 x+\sqrt{4 x^2+9}\right|+C\)
- D \(x \sqrt{4 x^2+9}+\frac{9}{4} \log \left|2 x+\sqrt{4 x^2+9}\right|+C\)
Answer & Solution
Correct Answer
(D) \(x \sqrt{4 x^2+9}+\frac{9}{4} \log \left|2 x+\sqrt{4 x^2+9}\right|+C\)
Step-by-step Solution
Detailed explanation
\( \int \sqrt{4 x^2+9} d x = \int \sqrt{(2x)^2+3^2} d x \) Let \(u=2x\), \(du=2dx \implies dx=\frac{1}{2}du \) \( \frac{1}{2} \int \sqrt{u^2+3^2} d u = \frac{1}{2} \left( \frac{u}{2}\sqrt{u^2+3^2} + \frac{3^2}{2}\log|u+\sqrt{u^2+3^2}| \right) + C \)…
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